Question

if $y = 9x + \frac{8}{x^2}$, find $\frac{dy}{dx}$ at $x = 1$

the value of $\frac{dy}{dx}$ at $x = 1$ is \\(\square\\) (simplify your answer )

Explanation:

Step1: Rewrite the function

$y = 9x + 8x^{-2}$

Step2: Differentiate term by term

$\frac{dy}{dx} = 9 + 8(-2)x^{-3} = 9 - 16x^{-3}$

Step3: Substitute x=1

$\frac{dy}{dx}\bigg|_{x=1} = 9 - 16(1)^{-3} = 9 - 16$

Answer:

-7